MBBS and BDS entrance Physics: NEET-UG questions on Nepal's MBBS syllabus: Modern Physics

2013 · NEET-UG (India) · 4 marks

In a n-type semiconductor, which of the following statements is true?
(a) Electrons are majority carriers and trivalent atoms are dopants (b) Electrons are minority carriers and pentavalent atoms are dopants (c) Holes are minority carriers and pentavalent atoms are dopants (d) Holes are majority carriers and trivalent atoms are dopants

Similar questions: p-type semiconductor (2019 · NEET-UG (India))

Answer

(c) Holes are minority carriers and pentavalent atoms are dopants

An n-type semiconductor is doped with pentavalent atoms (P, As, Sb), which donate free electrons. Electrons are therefore the majority carriers and holes the minority carriers. Options with trivalent dopants describe p-type material, and (b) wrongly calls electrons the minority.

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2019 · NEET-UG (India) · 4 marks

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is (nearly): ( $m_e=9\times10^{-31}$ kg)
(a) $12.2\times10^{-13}$ m (b) $12.2\times10^{-12}$ m (c) $12.2\times10^{-14}$ m (d) 12.2 nm

Similar questions: Accelerating potential from de Broglie wavelength (2020 · NEET-UG (India))

Answer

(b) $12.2\times10^{-12}$ m

$\lambda = \dfrac{h}{\sqrt{2meV}} \approx \dfrac{12.27}{\sqrt{V}}\ \text{\AA} = \dfrac{12.27}{100}\ \text{\AA} = 0.1227\ \text{\AA}$ . That is $1.22\times10^{-11}$ m $= 12.2\times10^{-12}$ m.

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2019 · NEET-UG (India) · 4 marks

For a p-type semiconductor, which of the following statements is true?
(a) Electrons are the majority carriers and trivalent atoms are the dopants. (b) Holes are the majority carriers and trivalent atoms are the dopants. (c) Holes are the majority carriers and pentavalent atoms are the dopants. (d) Electrons are the majority carriers and pentavalent atoms are the dopants.

Similar questions: n-type semiconductor (2013 · NEET-UG (India))

Answer

(b) Holes are the majority carriers and trivalent atoms are the dopants.

A p-type semiconductor is made by doping Si or Ge with trivalent atoms (B, Al, In). Each dopant creates a vacancy (hole) in the bonding. Holes are therefore the majority carriers. Pentavalent dopants give n-type, with electrons as majority carriers.

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2020 · NEET-UG (India) · 4 marks

An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227\times10^{-2}$ nm, the potential difference is
(a) $10^3$ V (b) $10^4$ V (c) 10 V (d) $10^2$ V

Similar questions: De Broglie wavelength of accelerated electron (2019 · NEET-UG (India))

Answer

(b) $10^4$ V

$\lambda=\dfrac{1.227}{\sqrt V}$ nm. Setting this to $1.227\times10^{-2}$ nm gives $\sqrt V=100$ . So $V=10^4$ V.

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2013 · NEET-UG (India) · 4 marks

Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is:
(a) $\frac{5}{27}$ (b) $\frac{3}{23}$ (c) $\frac{7}{29}$ (d) $\frac{9}{31}$

Answer

(a) $\frac{5}{27}$

Longest Lyman line ( $2\to1$ ): $\frac1{\lambda_L}=R\left(1-\frac14\right)=\frac{3R}{4}$ . Longest Balmer line ( $3\to2$ ): $\frac1{\lambda_B}=R\left(\frac14-\frac19\right)=\frac{5R}{36}$ . Ratio $\frac{\lambda_L}{\lambda_B}=\frac{4/3}{36/5}=\frac{5}{27}$ .

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2013 · NEET-UG (India) · 4 marks

The half-life of a radioactive isotope $X$ is 20 years. It decays to another element $Y$ which is stable. The two elements $X$ and $Y$ were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rock is estimated to be:
(a) 40 years (b) 60 years (c) 80 years (d) 100 years

Answer

(b) 60 years

X : Y = 1 : 7 means $\frac18$ of the original X is left, since Y is all from decayed X. $\frac18=\left(\frac12\right)^3$ , so 3 half-lives have passed: $3\times20=60$ years.

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2013 · NEET-UG (India) · 4 marks

A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 u. The energy liberated per u is: (Given 1 u = 931 MeV)
(a) 2.67 MeV (b) 26.7 MeV (c) 6.675 MeV (d) 13.35 MeV

Answer

(b) 26.7 MeV

Energy released $=\Delta m\times931=0.02866\times931\approx26.7$ MeV, which is the energy of the fusion of 4 H into one He (mass defect 0.02866 u). The question is loosely worded, and NTA's key takes the energy for this mass defect: 26.7 MeV. Option (a) wrongly divides by 10.

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2013 · NEET-UG (India) · 4 marks

For photoelectric emission from certain metal the cutoff frequency is $\nu$ . If radiation of frequency $2\nu$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be ( $m$ is the electron mass):
(a) $\sqrt{h\nu/(2m)}$ (b) $\sqrt{h\nu/m}$ (c) $\sqrt{2h\nu/m}$ (d) $2\sqrt{h\nu/m}$

Answer

(c) $\sqrt{2h\nu/m}$

Einstein's equation: $K_{max}=h(2\nu)-h\nu=h\nu$ . So $\frac12mv^2=h\nu$ , giving $v_{max}=\sqrt{\frac{2h\nu}{m}}$ .

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2013 · NEET-UG (India) · 4 marks

The wavelength $\lambda_e$ of an electron and $\lambda_p$ of a photon of same energy $E$ are related by:
(a) $\lambda_p\propto\lambda_e^2$ (b) $\lambda_p\propto\lambda_e$ (c) $\lambda_p\propto\sqrt{\lambda_e}$ (d) $\lambda_p\propto\frac{1}{\sqrt{\lambda_e}}$

Answer

(a) $\lambda_p\propto\lambda_e^2$

For the photon $\lambda_p=\frac{hc}{E}$ . For the electron $\lambda_e=\frac{h}{\sqrt{2mE}}$ , so $E=\frac{h^2}{2m\lambda_e^2}$ . Substituting: $\lambda_p=\frac{2mc}{h}\lambda_e^2$ , so $\lambda_p\propto\lambda_e^2$ .

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2013 · NEET-UG (India) · 4 marks

A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct?
(a) Diffraction pattern is not observed on the screen in the case of electrons (b) The angular width of the central maximum of the diffraction pattern will increase (c) The angular width of the central maximum will decrease (d) The angular width of the central maximum will be unaffected

Answer

(c) The angular width of the central maximum will decrease

Increasing speed increases momentum, so the de Broglie wavelength $\lambda=\frac hp$ decreases. The central maximum has angular half-width $\sin\theta=\frac{\lambda}{a}$ , so it narrows. Electrons do show diffraction, so (a) is wrong.

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2013 · NEET-UG (India) · 4 marks

In a common emitter (CE) amplifier having a voltage gain $G$ , the transistor used has transconductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will be:
(a) $\frac23G$ (b) $1.5G$ (c) $\frac13G$ (d) $\frac54G$

Answer

(a) $\frac23G$

Voltage gain of a CE amplifier is $A_v=\beta\frac{R_L}{R_i}=g_mR_L$ , since $g_m=\frac{\beta}{R_i}$ . With the same load, gain is proportional to $g_m$ . New gain $=G\times\frac{0.02}{0.03}=\frac23G$ . The current gain figures are a distraction.

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2014 · NEET-UG (India) · 4 marks

The barrier potential of a p-n junction depends on:
(a) type of semi conductor material (b) amount of doping (c) temperature
Which one of the following is correct?
(1) (b) only (2) (b) and (c) only (3) (a), (b) and (c) (4) (a) and (b) only

Answer

(3) (a), (b) and (c)

The barrier potential arises from the diffusion of carriers across the junction, which depends on the semiconductor material (band gap: about $0.7$ V for Si, $0.3$ V for Ge). It also depends on doping, since heavier doping gives a larger built-in potential, and on temperature, since it falls by about $2$ mV per $^\circ$ C. All three factors matter, so options that list only one or two are incomplete.

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2014 · NEET-UG (India) · 4 marks

The binding energy per nucleon of $^{7}_{3}\mathrm{Li}$ and $^{4}_{2}\mathrm{He}$ nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction $^{7}_{3}\mathrm{Li}+{}^{1}_{1}\mathrm{H}\rightarrow{}^{4}_{2}\mathrm{He}+{}^{4}_{2}\mathrm{He}+Q$ , the value of energy $Q$ released is:
(a) $-2.4$ MeV (b) 8.4 MeV (c) 17.3 MeV (d) 19.6 MeV

Answer

(c) 17.3 MeV

Energy released = total binding energy of the products minus that of the reactants (a bare proton has none). $Q=2\times(4\times7.06)-7\times5.60=56.48-39.20=17.28\approx17.3$ MeV. The negative value in (a) would mean energy is absorbed, but the products are more tightly bound, so energy is released.

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2014 · NEET-UG (India) · 4 marks

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is:
(a) 25 (b) 75 (c) 60 (d) 50

Answer

(b) 75

$\lambda=\frac{h}{\sqrt{2mK}}\propto K^{-1/2}$ . When $K$ becomes $16K$ , $\lambda'=\frac{\lambda}{\sqrt{16}}=\frac{\lambda}{4}$ . The change is $\frac{\lambda-\lambda/4}{\lambda}\times100=75\%$ (a decrease). Option (a) would be the new value as a percentage of the old one, not the change.

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2014 · NEET-UG (India) · 4 marks

Hydrogen atom in ground state is excited by a monochromatic radiation of $\lambda=975$ Å. Number of spectral lines in the resulting spectrum emitted will be:
(a) 2 (b) 6 (c) 10 (d) 3

Answer

(b) 6

Photon energy $E=\frac{12375}{975}\approx12.75$ eV. This equals $E_4-E_1=13.6\left(1-\frac1{16}\right)=12.75$ eV, so the atom is excited to $n=4$ . From $n=4$ the number of possible spectral lines is $\frac{n(n-1)}{2}=6$ . Option (d), $3$ , would count only transitions down to the ground state.

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2014 · NEET-UG (India) · 4 marks

A radio isotope $X$ with a half life $1.4\times10^{9}$ years decays to $Y$ which is stable. A sample of the rock from a cave was found to contain $X$ and $Y$ in the ratio 1 : 7. The age of the rock is:
(a) $3.92\times10^{9}$ years (b) $4.20\times10^{9}$ years (c) $8.40\times10^{9}$ years (d) $1.96\times10^{9}$ years

Answer

(b) $4.20\times10^{9}$ years

$X:Y=1:7$ means $X$ is $\frac18$ of the original amount, so $\frac{N}{N_0}=\frac18=\left(\frac12\right)^3$ , i.e. $3$ half-lives. Age $=3\times1.4\times10^9=4.2\times10^9$ years. Option (c) uses $6$ half-lives, and (a) assumes a different number of half-lives; only $n=3$ fits the ratio.

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2014 · NEET-UG (India) · 4 marks

When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is:
(a) 1.0 eV (b) 1.3 eV (c) 1.5 eV (d) 0.65 eV

Answer

(a) 1.0 eV

Let the initial photon energy be $E$ : $E-W=0.5$ and $1.2E-W=0.8$ . Subtracting, $0.2E=0.3$ , so $E=1.5$ eV. Then $W=1.5-0.5=1.0$ eV. Option (c) is the photon energy, not the work function.

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2015 · NEET-UG (India) · 4 marks

A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda}{2}$ . If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is: ( $h$ = Planck's constant, $c$ = speed of light)
(a) $\frac{2hc}{\lambda}$ (b) $\frac{hc}{3\lambda}$ (c) $\frac{hc}{2\lambda}$ (d) $\frac{hc}{\lambda}$

Answer

(c) $\frac{hc}{2\lambda}$

Einstein's equation: $K_1=\frac{hc}{\lambda}-W$ and $K_2=\frac{2hc}{\lambda}-W$ , with $K_2=3K_1$ . $\frac{2hc}{\lambda}-W=\frac{3hc}{\lambda}-3W$ , so $2W=\frac{hc}{\lambda}$ , giving $W=\frac{hc}{2\lambda}$ . Option (d), $W=\frac{hc}{\lambda}$ , would make $K_1=0$ , so $K_2=3K_1$ would not hold with $K_2\neq0$ .

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2015 · NEET-UG (India) · 4 marks

The input signal given to a CE amplifier having a voltage gain of 150 is $V_i = 2\cos\left(15t+\frac{\pi}{3}\right)$ . The corresponding output signal will be:
(a) $2\cos\left(15t+\frac{5\pi}{6}\right)$ (b) $300\cos\left(15t+\frac{4\pi}{3}\right)$ (c) $300\cos\left(15t+\frac{\pi}{3}\right)$ (d) $75\cos\left(15t+\frac{2\pi}{3}\right)$

Answer

(b) $300\cos\left(15t+\frac{4\pi}{3}\right)$

A CE amplifier amplifies the amplitude by the gain: $2\times150=300$ . It also introduces a phase shift of $\pi$ between input and output, so the phase is $\frac{\pi}{3}+\pi=\frac{4\pi}{3}$ . Option (c) has the right amplitude but no phase reversal, which would be true of a CB or CC amplifier.

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2015 · NEET-UG (India) · 4 marks

In the given figure, a diode D is connected in series with an external resistance $R=100\ \Omega$ and an e.m.f. of 3.5 V (the diode is forward biased). If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be:
(a) 20 mA (b) 35 mA (c) 30 mA (d) 40 mA

Answer

(c) 30 mA

The diode is forward biased and has a barrier potential of 0.5 V, so the voltage across the resistor is $3.5-0.5=3.0$ V. Then $I = \dfrac{3.0}{100}=0.03$ A $=30$ mA.

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2015 · NEET-UG (India) · 4 marks

Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of the emitted electron is:
(a) $\geq 2.8\times10^{-9}$ m (b) $\leq 2.8\times10^{-12}$ m (c) $< 2.8\times10^{-10}$ m (d) $< 2.8\times10^{-9}$ m

Answer

(a) $\geq 2.8\times10^{-9}$ m

Photon energy $=1240/500=2.48$ eV, so the maximum kinetic energy is $K_{max}=2.48-2.28=0.20$ eV. The de Broglie wavelength is $\lambda=\dfrac{h}{\sqrt{2mK}}=\dfrac{12.27}{\sqrt{0.20}}\ \text{Å}\approx 27.4\ \text{Å}\approx 2.8\times10^{-9}$ m. Emitted electrons have kinetic energy up to this maximum, so their wavelengths are at least this value, i.e. $\geq 2.8\times10^{-9}$ m.

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2015 · NEET-UG (India) · 4 marks

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:
(a) $\frac{27}{5}$ (b) $\frac{5}{27}$ (c) $\frac{4}{9}$ (d) $\frac{9}{4}$

Answer

(b) $\dfrac{5}{27}$

Longest wavelength in the Lyman series (2 to 1): $\dfrac1{\lambda_L}=R\left(1-\dfrac14\right)=\dfrac{3R}{4}$ . Longest in the Balmer series (3 to 2): $\dfrac1{\lambda_B}=R\left(\dfrac14-\dfrac19\right)=\dfrac{5R}{36}$ . So $\dfrac{\lambda_L}{\lambda_B}=\dfrac{5/36}{3/4}=\dfrac{5}{27}$ . The inverse, $27/5$ , is the tempting wrong option.

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2016 · NEET-UG (India) · 4 marks

When a metallic surface is illuminated with radiation of wavelength $\lambda$ , the stopping potential is $V$ . If the same surface is illuminated with radiation of wavelength $2\lambda$ , the stopping potential is $\frac{V}{4}$ . The threshold wavelength for the metallic surface is:
(a) $3\lambda$ (b) $4\lambda$ (c) $5\lambda$ (d) $\frac{5}{2}\lambda$

Answer

(a) $3\lambda$

Einstein's equation: $\dfrac{hc}{\lambda}=\phi+eV$ and $\dfrac{hc}{2\lambda}=\phi+\dfrac{eV}{4}$ . Let $E=hc/\lambda$ . Subtracting: $\dfrac E2=\dfrac{3eV}4$ , so $eV=\dfrac{2E}{3}$ and $\phi=E-\dfrac{2E}3=\dfrac E3$ . Threshold wavelength $\lambda_0=\dfrac{hc}{\phi}=3\lambda$ .

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2016 · NEET-UG (India) · 4 marks

When an $\alpha$ -particle of mass $m$ moving with velocity $v$ bombards on a heavy nucleus of charge $Ze$ , its distance of closest approach from the nucleus depends on $m$ as:
(a) $m$ (b) $\frac{1}{m}$ (c) $\frac{1}{\sqrt m}$ (d) $\frac{1}{m^2}$

Answer

(b) $\dfrac1m$

At closest approach the kinetic energy turns entirely into electrostatic potential energy: $\tfrac12mv^2=\dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{r_0}$ . So $r_0=\dfrac{Ze^2}{\pi\varepsilon_0mv^2}\propto\dfrac1m$ for a given velocity.

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2016 · NEET-UG (India) · 4 marks

Consider the junction diode as ideal. A diode and a $1\ \mathrm{k\Omega}$ resistor are in series between points A and B, with the diode's anode towards A. A is at $+4$ V and B is at $-6$ V. The value of current flowing through AB is:
(a) $10^{-3}$ A (b) 0 A (c) $10^{-2}$ A (d) $10^{-1}$ A

Answer

(c) $10^{-2}$ A

The anode is at $+4$ V and the cathode side is at $-6$ V, so the ideal diode is forward biased and acts as a short with zero drop. Then $I=\dfrac{4-(-6)}{1000}=10^{-2}$ A. Zero current (option b) would apply only if the diode were reverse biased.

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2016 · NEET-UG (India) · 4 marks

Given the value of Rydberg constant is $10^{7}\ \mathrm{m^{-1}}$ , the wave number of the last line of the Balmer series in hydrogen spectrum will be:
(a) $2.5\times10^{7}\ \mathrm{m^{-1}}$ (b) $0.025\times10^{4}\ \mathrm{m^{-1}}$ (c) $0.5\times10^{7}\ \mathrm{m^{-1}}$ (d) $0.25\times10^{7}\ \mathrm{m^{-1}}$

Answer

(d) $0.25\times10^{7}\ \mathrm{m^{-1}}$

For the Balmer series, $\bar\nu=R\left(\frac1{2^2}-\frac1{n^2}\right)$ . The last (series limit) line has $n\to\infty$ , so $\bar\nu=\dfrac R4=\dfrac{10^7}{4}=0.25\times10^7\ \mathrm{m^{-1}}$ .

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2016 · NEET-UG (India) · 4 marks

An electron of mass $m$ and a photon have same energy $E$ . The ratio of de-Broglie wavelengths associated with them is: ( $c$ being velocity of light)
(a) $\frac1c\left(\frac{2m}{E}\right)^{1/2}$ (b) $\frac1c\left(\frac{E}{2m}\right)^{1/2}$ (c) $\left(\frac{E}{2m}\right)^{1/2}$ (d) $c\,(2mE)^{1/2}$

Answer

(b) $\dfrac1c\left(\dfrac{E}{2m}\right)^{1/2}$

For the electron, $E=\dfrac{p^2}{2m}$ , so $\lambda_e=\dfrac h{\sqrt{2mE}}$ . For the photon, $E=pc$ , so $\lambda_{ph}=\dfrac{hc}{E}$ . The ratio is $\dfrac{\lambda_e}{\lambda_{ph}}=\dfrac{E}{c\sqrt{2mE}}=\dfrac1c\sqrt{\dfrac{E}{2m}}$ .

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2016 · NEET-UG (India) · 4 marks

An npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of $800\ \Omega$ is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is $192\ \Omega$ , the voltage gain and the power gain of the amplifier will respectively be:
(a) 4, 3.69 (b) 4, 3.84 (c) 3.69, 3.84 (d) 4, 4

Answer

(b) 4, 3.84

Collector current is $I_c=\dfrac{0.8}{800}=1$ mA. Voltage gain $A_V=\beta\dfrac{R_L}{R_{in}}=0.96\times\dfrac{800}{192}=4$ . Power gain $=\beta A_V=0.96\times4=3.84$ . The value 3.69 does not follow from these relations.

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2017 · NEET-UG (India) · 4 marks

In a common emitter transistor amplifier the audio signal voltage across the collector is 3 V. The resistance of collector is $3\ \mathrm{k}\Omega$ . If current gain is 100 and the base resistance is $2\ \mathrm{k}\Omega$ , the voltage and power gain of the amplifier is
(a) 15 and 200 (b) 150 and 15000 (c) 20 and 2000 (d) 200 and 1000

Answer

(b) 150 and 15000

Voltage gain $A_v=\beta\,R_C/R_B = 100\times 3/2 = 150$ . Power gain is current gain times voltage gain: $A_p=\beta A_v = 100\times150=15000$ .

The given 3 V collector signal is not needed; only the resistance ratio and $\beta$ matter.

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2017 · NEET-UG (India) · 4 marks

The de Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (kelvin) and mass $m$ , is
(a) $\dfrac{h}{\sqrt{3mkT}}$ (b) $\dfrac{2h}{\sqrt{3mkT}}$ (c) $\dfrac{2h}{\sqrt{mkT}}$ (d) $\dfrac{h}{\sqrt{mkT}}$

Answer

(a) $\dfrac{h}{\sqrt{3mkT}}$

A thermal neutron has average kinetic energy $K=\tfrac32kT$ . Momentum is $p=\sqrt{2mK}=\sqrt{3mkT}$ .

So $\lambda=h/p=\dfrac{h}{\sqrt{3mkT}}$ .

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2017 · NEET-UG (India) · 4 marks

Radioactive material $A$ has decay constant $8\lambda$ and material $B$ has decay constant $\lambda$ . Initially they have the same number of nuclei. After what time will the ratio of number of nuclei of material $B$ to that of $A$ be $\frac{1}{e}$ ?
(a) $\dfrac{1}{7\lambda}$ (b) $\dfrac{1}{8\lambda}$ (c) $\dfrac{1}{9\lambda}$ (d) $\dfrac{1}{\lambda}$

Answer

(a) $\dfrac{1}{7\lambda}$

The number of nuclei is $N=N_0e^{-\lambda t}$ , so the two samples differ by the factor $e^{-(8\lambda-\lambda)t}=e^{-7\lambda t}$ (the faster-decaying material over the slower one, which is how the official key reads this question).

Setting $e^{-7\lambda t}=1/e$ gives $7\lambda t=1$ , so $t=1/(7\lambda)$ .

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2017 · NEET-UG (India) · 4 marks

The photoelectric threshold wavelength of silver is $3250\times10^{-10}$ m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536\times10^{-10}$ m is (Given $h=4.14\times10^{-15}$ eV s and $c=3\times10^{8}\ \mathrm{m\,s^{-1}}$ )
(a) $\approx0.6\times10^{6}\ \mathrm{m\,s^{-1}}$ (b) $\approx61\times10^{3}\ \mathrm{m\,s^{-1}}$ (c) $\approx0.3\times10^{6}\ \mathrm{m\,s^{-1}}$ (d) $\approx6\times10^{5}\ \mathrm{m\,s^{-1}}$

Answer

(a) $\approx0.6\times10^{6}\ \mathrm{m\,s^{-1}}$

$hc=4.14\times10^{-15}\times3\times10^8=1242$ eV nm. $K_{max}=hc\left(\dfrac1{253.6}-\dfrac1{325}\right)\approx4.90-3.82=1.07$ eV.

$v=\sqrt{2K/m}=\sqrt{2\times1.07\times1.6\times10^{-19}/9.1\times10^{-31}}\approx6\times10^5\ \mathrm{m/s}=0.6\times10^6\ \mathrm{m/s}$ . Option (d) states the same value; the official key marks the $0.6\times10^6$ form. The $0.3\times10^6$ option would need a quarter of the energy.

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2017 · NEET-UG (India) · 4 marks

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is
(a) 1 (b) 4 (c) 0.5 (d) 2

Answer

(b) 4

The last line of a series is the series limit ( $n\to\infty$ ). Balmer: $\dfrac1\lambda=R\left(\dfrac1{2^2}-0\right)=\dfrac R4$ . Lyman: $\dfrac1\lambda=R\left(\dfrac1{1^2}\right)=R$ .

Ratio $\lambda_B/\lambda_L=4$ .

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2018 · NEET-UG (India) · 4 marks

In a p-n junction diode, change in temperature due to heating
(a) affects the overall V-I characteristics of p-n junction (b) affects only reverse resistance (c) does not affect resistance of p-n junction (d) affects only forward resistance

Answer

(a) affects the overall V-I characteristics of p-n junction

Heating generates more electron-hole pairs, raising the minority-carrier concentration and the reverse saturation current, and it also shifts the junction (knee) voltage. Both forward and reverse behaviour change, so the whole V-I characteristic is affected, not just one resistance.

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2018 · NEET-UG (India) · 4 marks

When light of frequency $2\nu_0$ (where $\nu_0$ is the threshold frequency) is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$ . When the frequency of the incident radiation is increased to $5\nu_0$ , the maximum velocity of electrons emitted from the same plate is $v_2$ . The ratio of $v_1$ to $v_2$ is
(a) $2:1$ (b) $1:2$ (c) $4:1$ (d) $1:4$

Answer

(b) $1:2$

Einstein: $K_{max} = h\nu - h\nu_0$ . For $2\nu_0$ , $K_1 = h\nu_0$ ; for $5\nu_0$ , $K_2 = 4h\nu_0$ . Since $v \propto \sqrt{K_{max}}$ , $v_1/v_2 = \sqrt{1/4} = 1/2$ . Speed is not proportional to frequency, so $2:5$ -style guesses fail.

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2018 · NEET-UG (India) · 4 marks

For a radioactive material, half-life is 10 minutes. If initially there are 600 nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
(a) 15 (b) 20 (c) 30 (d) 10

Answer

(b) 20

If 450 of 600 nuclei disintegrate, 150 remain, i.e. $N/N_0 = 1/4 = (1/2)^2$ . That is two half-lives, so $t = 2 \times 10 = 20$ min. Option (a) 15 min would wrongly treat decay as linear.

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2018 · NEET-UG (India) · 4 marks

An electron of mass $m$ with an initial velocity $\vec V = V_0\hat i$ ( $V_0>0$ ) enters an electric field $\vec E = -E_0\hat i$ ( $E_0$ = constant $>0$ ) at $t=0$ . If $\lambda_0$ is its de Broglie wavelength initially, then its de Broglie wavelength at time $t$ is
(a) $\lambda_0$ (b) $\dfrac{\lambda_0}{1+\dfrac{eE_0}{mV_0}t}$ (c) $\lambda_0 t$ (d) $\lambda_0\left(1+\dfrac{eE_0}{mV_0}t\right)$

Answer

(b) $\dfrac{\lambda_0}{1+\dfrac{eE_0}{mV_0}t}$

The field is along $-\hat i$ , so the force on the electron is $+eE_0\hat i$ and it speeds up: $V = V_0 + \dfrac{eE_0}{m}t$ . $\lambda = h/(mV)$ , so $\lambda = \lambda_0 V_0/V = \lambda_0/\left(1+\dfrac{eE_0 t}{mV_0}\right)$ . Wavelength decreases as speed grows, ruling out (d).

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2018 · NEET-UG (India) · 4 marks

The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is
(a) $1:-2$ (b) $1:1$ (c) $2:-1$ (d) $1:-1$

Answer

(d) $1:-1$

In a Bohr orbit $K = -E$ (total energy $E = -K$ , potential energy $U = -2K$ ). So $K:E = 1:-1$ .

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2019 · NEET-UG (India) · 4 marks

The total energy of an electron in an atom in an orbit is $-3.4$ eV. Its kinetic and potential energies are, respectively
(a) $-3.4$ eV, $-3.4$ eV (b) $-3.4$ eV, $-6.8$ eV (c) $3.4$ eV, $-6.8$ eV (d) $3.4$ eV, $3.4$ eV

Answer

(c) $3.4$ eV, $-6.8$ eV

In a bound orbit, $K=-E$ and $U=2E$ . With $E=-3.4$ eV, $K=+3.4$ eV and $U=-6.8$ eV, and $K+U=-3.4$ eV checks. A negative kinetic energy is impossible.

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2019 · NEET-UG (India) · 4 marks

$\alpha$ -particle consists of
(a) 2 protons and 2 neutrons only (b) 2 electrons, 2 protons and 2 neutrons (c) 2 electrons and 4 protons only (d) 2 protons only

Answer

(a) 2 protons and 2 neutrons only

An alpha particle is a helium nucleus, ${}^4_2\mathrm{He}$ : 2 protons and 2 neutrons, charge $+2e$ . It has no electrons, so the options with electrons are wrong, and 2 protons alone would give mass number 2.

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2020 · NEET-UG (India) · 4 marks

Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?
(a) one-fourth (b) zero (c) doubled (d) four times

Answer

(b) zero

The threshold frequency is $\nu_0$ ; the new frequency is $0.75\nu_0<\nu_0$ . Below threshold no photoelectrons are emitted whatever the intensity, so photocurrent is zero. Doubling intensity only raises the number of photons, not their energy.

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2020 · NEET-UG (India) · 4 marks

For transistor action, which of the following statements is correct?
(a) Both emitter junction as well as the collector junction are forward biased. (b) The base region must be very thin and lightly doped. (c) Base, emitter and collector regions should have the same doping concentrations. (d) Base, emitter and collector regions should have the same size.

Answer

(b) The base region must be very thin and lightly doped.

A thin, lightly doped base lets most carriers from the emitter cross to the collector instead of recombining. The emitter junction is forward biased and the collector junction reverse biased, so (a) is wrong. The emitter is heavily doped, the base light and the collector moderate, and the regions differ in size.

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2020 · NEET-UG (India) · 4 marks

The increase in the width of the depletion region in a p-n junction diode is due to
(a) both forward bias and reverse bias (b) increase in forward current (c) forward bias only (d) reverse bias only

Answer

(d) reverse bias only

Reverse bias pulls majority carriers away from the junction, so the depletion region widens. Forward bias narrows it, and a larger forward current only means it is narrower still.

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2021 · NEET-UG (India) · 4 marks

A radioactive nucleus $^{A}_{Z}X$ undergoes spontaneous decay in the sequence $^{A}_{Z}X \rightarrow {}_{Z-1}B \rightarrow {}_{Z-3}C \rightarrow {}_{Z-2}D$ , where $Z$ is the atomic number of element X. The possible decay particles in the sequence are:
(a) $\beta^+, \alpha, \beta^-$ (b) $\beta^-, \alpha, \beta^+$ (c) $\alpha, \beta^-, \beta^+$ (d) $\alpha, \beta^+, \beta^-$

Answer

(a) $\beta^+, \alpha, \beta^-$

$Z\to Z-1$ is $\beta^+$ decay (a proton becomes a neutron). $Z-1\to Z-3$ loses 2 units of charge, so it is $\alpha$ emission. $Z-3\to Z-2$ gains one unit of charge, so it is $\beta^-$ . Options (b), (c) and (d) put $\beta^-$ or $\alpha$ first, which would raise $Z$ or lower it by 2 at the first step.

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2021 · NEET-UG (India) · 4 marks

The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.
(a) Current in n-type > current in p-type (b) No current will flow in p-type, current will only flow in n-type (c) Current in n-type = current in p-type (d) Current in p-type > current in n-type

Answer

(a) Current in n-type > current in p-type

Current $I=nevA$ with drift velocity $v=\mu E$ . For equal carrier concentrations and the same field, the current depends on mobility. Electrons are more mobile than holes ( $\mu_e>\mu_h$ ), so the n-type current is larger. Both types conduct, so (b) is wrong.

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2021 · NEET-UG (India) · 4 marks

The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of $3.3 \times 10^{-3}$ watt will be: ( $h = 6.6 \times 10^{-34}$ Js)
(a) $10^{16}$ (b) $10^{15}$ (c) $10^{18}$ (d) $10^{17}$

Answer

(a) $10^{16}$

Energy per photon $=\frac{hc}{\lambda}=\frac{6.6\times10^{-34}\times3\times10^8}{6\times10^{-7}}=3.3\times10^{-19}$ J. Photons per second $=\frac{P}{E}=\frac{3.3\times10^{-3}}{3.3\times10^{-19}}=10^{16}$ .

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2021 · NEET-UG (India) · 4 marks

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If a photoelectron of mass $m$ emitted from the surface has de-Broglie wavelength $\lambda_d$ , then:
(a) $\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2$ (b) $\lambda = \left(\frac{2h}{mc}\right)\lambda_d^2$ (c) $\lambda = \left(\frac{2m}{hc}\right)\lambda_d^2$ (d) $\lambda_d = \left(\frac{2mc}{h}\right)\lambda^2$

Answer

(a) $\lambda=\left(\frac{2mc}{h}\right)\lambda_d^2$

With negligible work function, the photon energy becomes kinetic energy: $\frac{hc}{\lambda}=\frac{p^2}{2m}$ . Using $p=\frac{h}{\lambda_d}$ , $\frac{hc}{\lambda}=\frac{h^2}{2m\lambda_d^2}$ , so $\lambda=\frac{2mc}{h}\lambda_d^2$ .

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2021 · NEET-UG (India) · 4 marks

A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the binding energy in the process is:
(a) 804 MeV (b) 216 MeV (c) 0.9 MeV (d) 9.4 MeV

Answer

(b) 216 MeV

Binding energy of the fragments $=2\times120\times8.5=2040$ MeV. Binding energy of the parent $=240\times7.6=1824$ MeV. Gain $=2040-1824=216$ MeV. The value 0.9 MeV is only the per-nucleon difference.

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2021 · NEET-UG (India) · 4 marks

Consider the following statements (A) and (B) and identify the correct answer.
A. A zener diode is connected in reverse bias, when used as a voltage regulator. B. The potential barrier of p-n junction lies between 0.1 V to 0.3 V.
(a) (A) is correct (B) is incorrect. (b) (A) is incorrect but (B) is correct. (c) (A) and (B) both are correct. (d) (A) and (B) both are incorrect.

Answer

(a) (A) is correct (B) is incorrect.

A zener diode is operated in reverse bias (breakdown region) to regulate voltage, so A is correct. The potential barrier of a p-n junction is about 0.7 V for silicon and 0.3 V for germanium, not 0.1 to 0.3 V, so B is incorrect.

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2021 · NEET-UG (India) · 4 marks

A half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be:
(a) $\frac{2}{3}$ (b) $\frac{2}{3\sqrt{2}}$ (c) $\frac{1}{2}$ (d) $\frac{1}{2\sqrt{2}}$

Answer

(d) $\frac{1}{2\sqrt2}$

$\frac{A}{A_0}=\left(\frac{1}{2}\right)^{t/T_{1/2}}=\left(\frac12\right)^{150/100}=\left(\frac12\right)^{3/2}=\frac{1}{2\sqrt2}$ . Option (c) would be the fraction after exactly one half-life.

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2022 · NEET-UG (India) · 4 marks

Let $T_1$ and $T_2$ be the energy of an electron in the first and second excited states of hydrogen atom, respectively. According to the Bohr's model of an atom, the ratio $T_1 : T_2$ is:
(a) 1 : 4 (b) 4 : 1 (c) 4 : 9 (d) 9 : 4

Answer

(d) 9 : 4

In the Bohr model $E_n\propto\frac{1}{n^2}$ . The first excited state is $n=2$ and the second is $n=3$ . So $T_1:T_2=\frac14:\frac19=9:4$ . Option (c) 4 : 9 inverts the ratio.

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2022 · NEET-UG (India) · 4 marks

In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be:
(a) Zero (b) 30 Hz (c) 60 Hz (d) 120 Hz

Answer

(c) 60 Hz

A half wave rectifier passes only one half of each input cycle, so there is one output pulse per input cycle and the output frequency equals the input frequency. A full wave rectifier would give 120 Hz, which is the tempting wrong option.

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2022 · NEET-UG (India) · 4 marks

In the given nuclear reaction, the element X is: $^{22}_{11}\mathrm{Na} \rightarrow X + e^{+} + \nu$
(a) $^{23}_{11}\mathrm{Na}$ (b) $^{23}_{10}\mathrm{Ne}$ (c) $^{22}_{10}\mathrm{Ne}$ (d) $^{22}_{12}\mathrm{Mg}$

Answer

(c) $^{22}_{10}\mathrm{Ne}$

In positron emission a proton changes into a neutron, so the atomic number falls by 1 and the mass number is unchanged: $^{22}_{11}\mathrm{Na}\to{}^{22}_{10}\mathrm{Ne}+e^++\nu$ . Charge and mass number balance.

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2022 · NEET-UG (India) · 4 marks

A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64. The ratio of radius of two daughter nuclei respectively is:
(a) 1 : 1 (b) 4 : 5 (c) 5 : 4 (d) 25 : 16

Answer

(c) 5 : 4

Nuclear radius $R = R_0A^{1/3}$ . So $R_1:R_2 = (125)^{1/3}:(64)^{1/3} = 5:4$ .

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2023 · NEET-UG (India) · 4 marks

The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to:
(a) $V$ (b) $\frac{1}{V}$ (c) $\frac{1}{\sqrt{V}}$ (d) $\sqrt{V}$

Answer

(b) $\dfrac{1}{V}$

All the kinetic energy $eV$ is converted into one photon at the cut-off: $eV = hc/\lambda_{min}$ , so $\lambda_{min} = \dfrac{hc}{eV}\propto \dfrac1V$ .

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2023 · NEET-UG (India) · 4 marks

Given below are two statements:
Statement I: Photovoltaic devices can convert optical radiation into electricity. Statement II: Zener diode is designed to operate under reverse bias in breakdown region.
In the light of the above statements, choose the most appropriate answer from the options given below:
(a) Both Statement I and Statement II are correct. (b) Both Statement I and Statement II are incorrect. (c) Statement I is correct but Statement II is incorrect. (d) Statement I is incorrect but Statement II is correct.

Answer

(a) Both Statement I and Statement II are correct.

Photovoltaic devices (solar cells) convert optical radiation into electricity. A Zener diode is designed to work in reverse bias in the breakdown region, which is why it is used as a voltage regulator. Both statements are correct.

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2023 · NEET-UG (India) · 4 marks

In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$ . The shortest wavelength in the Brackett series is:
(a) $2\lambda$ (b) $4\lambda$ (c) $9\lambda$ (d) $16\lambda$

Answer

(b) $4\lambda$

The shortest wavelength of a series corresponds to $n_2=\infty$ : $\dfrac1\lambda = \dfrac{R}{n_1^2}$ , so $\lambda\propto n_1^2$ . Balmer has $n_1=2$ and Brackett has $n_1=4$ , so $\lambda_{Br} = \dfrac{16}{4}\lambda = 4\lambda$ .

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2023 · NEET-UG (India) · 4 marks

The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14 eV, 2.30 eV and 2.75 eV respectively. If incident electromagnetic radiation has an incident energy of 2.20 eV, which of these photosensitive surfaces may emit photoelectrons?
(a) Cs only (b) Both Na and K (c) K only (d) Na only

Answer

(a) Cs only

A photoelectron is emitted only if the photon energy is at least the work function. The incident energy is 2.20 eV, which exceeds Cs ( $2.14$ eV) but is below K ( $2.30$ eV) and Na ( $2.75$ eV), so K and Na cannot emit.

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2023 · NEET-UG (India) · 4 marks

The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $\left(\frac{1}{16}\right)^{th}$ of its initial value?
(a) 20 minutes (b) 40 minutes (c) 60 minutes (d) 80 minutes

Answer

(d) 80 minutes

Activity falls to $\frac{1}{16}=\left(\frac12\right)^4$ , which needs 4 half-lives. Time $=4\times20=80$ minutes.

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2023 · NEET-UG (India) · 4 marks

A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a load resistance. Which of these components remove the ac ripple from the rectified output?
(a) A centre-tapped transformer (b) p-n junction diodes (c) Capacitor (d) Load resistance

Answer

(c) Capacitor

The rectifier diodes and centre-tapped transformer only produce a pulsating DC. The capacitor, connected across the load, charges at the peaks and discharges slowly in between, smoothing out the ac ripple.

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2023 · NEET-UG (India) · 4 marks

The radius of innermost orbit of hydrogen atom is $5.3 \times 10^{-11}$ m. What is the radius of third allowed orbit of hydrogen atom?
(a) 0.53 $\mathrm{\AA}$ (b) 1.06 $\mathrm{\AA}$ (c) 1.59 $\mathrm{\AA}$ (d) 4.77 $\mathrm{\AA}$

Answer

(d) $4.77\ \mathrm{\AA}$

Bohr radius scales as $r_n\propto n^2$ . For $n=3$ : $r_3=9\times5.3\times10^{-11}=4.77\times10^{-10}$ m $=4.77\ \mathrm{\AA}$ . Option (c) wrongly uses $3\times$ the radius.

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2024 · NEET-UG (India) · 4 marks

Given below are two statements:
Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges. Statement II: Atoms of each element are stable and emit their characteristic spectrum.
In the light of the above statements, choose the most appropriate answer from the options given below:
(a) Both Statement I and Statement II are incorrect. (b) Statement I is correct but Statement II is incorrect. (c) Statement I is incorrect but Statement II is correct (d) Both Statement I and Statement II are correct

Answer

(d) Both Statement I and Statement II are correct

An atom has equal numbers of protons and electrons, so it is electrically neutral. Each element is stable in its ground state and gives its own characteristic line spectrum when excited, so both statements are correct.

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2024 · NEET-UG (India) · 4 marks

Consider the following statements A and B and identify the correct answer:
A. For a solar cell, the I-V characteristic lies in the IV quadrant of the given graph. B. In a reverse biased pn junction diode, the current measured in ( $\mu$ A), is due to majority charge carriers.
(a) A is incorrect but B is correct (b) Both A and B are correct (c) Both A and B are incorrect (d) A is correct but B is incorrect.

Answer

(d) A is correct but B is incorrect.

A solar cell is operated as a generator, so its I-V curve lies in the fourth quadrant. In reverse bias the small microampere current is due to minority carriers, not majority carriers, so B is incorrect.

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2024 · NEET-UG (India) · 4 marks

If $c$ is the velocity of light in free space, the correct statements about photon among the following are:
A. The energy of a photon is $E = h\nu$ . B. The velocity of a photon is $c$ . C. The momentum of a photon, $p = \frac{h\nu}{c}$ . D. In a photon-electron collision, both total energy and total momentum are conserved. E. Photon possesses positive charge.
Choose the correct answer from the options given below:
(a) A, B, C and D only (b) A, C and D only (c) A, B, D and E only (d) A and B only

Answer

(a) A, B, C and D only

Photon energy is $h\nu$ , its speed is $c$ , and its momentum is $\frac{h\nu}{c}$ . In a photon-electron collision (Compton scattering), total energy and momentum are both conserved. A photon is neutral, so E is false.

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2024 · NEET-UG (India) · 4 marks

Match List I with List II.
List - I (Spectral lines of hydrogen for transitions from): A. $n_2 = 3$ to $n_1 = 2$ ; B. $n_2 = 4$ to $n_1 = 2$ ; C. $n_2 = 5$ to $n_1 = 2$ ; D. $n_2 = 6$ to $n_1 = 2$
List - II (Wavelengths, nm): I. 410.2; II. 434.1; III. 656.3; IV. 486.1
Choose the correct answer from the options given below:
(a) A-III, B-IV, C-II, D-I (b) A-IV, B-III, C-I, D-II (c) A-I, B-II, C-III, D-IV (d) A-II, B-I, C-IV, D-III

Answer

(a) A-III, B-IV, C-II, D-I

These are the Balmer lines ending at $n=2$ . The larger the jump, the shorter the wavelength: $3\to2$ is H-alpha (656.3 nm), $4\to2$ is 486.1 nm, $5\to2$ is 434.1 nm and $6\to2$ is 410.2 nm. Wavelength falls as $n_2$ rises, since $\frac{1}{\lambda}=R\left(\frac14-\frac1{n_2^2}\right)$ increases with $n_2$ . This rules out options (c) and (d), which pair the longest wavelength with a higher level.

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2025 · NEET-UG (India) · 4 marks

A particle of mass $m$ is moving around the origin with a constant force $F$ pulling it towards the origin. If Bohr model is used to describe its motion, the radius of the $n$ th orbit and the particle's speed $v$ in the orbit depend on $n$ as:
(a) $r \propto n^{2/3}$ ; $v \propto n^{1/3}$ (b) $r \propto n^{4/3}$ ; $v \propto n^{-1/3}$ (c) $r \propto n^{1/3}$ ; $v \propto n^{1/3}$ (d) $r \propto n^{1/3}$ ; $v \propto n^{2/3}$

Answer

(a) $r\propto n^{2/3}$ ; $v\propto n^{1/3}$

The constant force supplies the centripetal force: $F=\frac{mv^2}{r}$ . Bohr quantisation gives $mvr=\frac{nh}{2\pi}$ . Substituting $v=\frac{nh}{2\pi mr}$ gives $F=\frac{n^2h^2}{4\pi^2 m r^3}$ , so $r^3\propto n^2$ and $r\propto n^{2/3}$ . Then $v\propto\frac nr\propto n^{1-2/3}=n^{1/3}$ .

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2025 · NEET-UG (India) · 4 marks

De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius = 0.052 nm):
(a) 1.67 nm (b) 2.67 nm (c) 0.067 nm (d) 0.67 nm

Answer

(d) 0.67 nm

For the $n$ th Bohr orbit, $2\pi r_n=n\lambda$ , so $\lambda=\frac{2\pi r_n}{n}$ with $r_n=n^2a_0$ . $r_2=4\times0.052=0.208$ nm, and $\lambda=\frac{2\pi\times0.208}{2}\approx0.653$ nm, which is closest to 0.67 nm.

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2025 · NEET-UG (India) · 4 marks

A photon and an electron (mass $m$ ) have the same energy $E$ . The ratio (photon/electron) of their de Broglie wavelengths is: ( $c$ is the speed of light)
(a) $c\sqrt{\frac{2m}{E}}$ (b) $\frac{1}{c}\sqrt{\frac{E}{2m}}$ (c) $\sqrt{\frac{E}{2m}}$ (d) $c\sqrt{2mE}$

Answer

(a) $c\sqrt{\frac{2m}{E}}$

Photon: $E=\frac{hc}{\lambda_p}$ , so $\lambda_p=\frac{hc}{E}$ . Electron: $p=\sqrt{2mE}$ , so $\lambda_e=\frac{h}{\sqrt{2mE}}$ . $\frac{\lambda_p}{\lambda_e}=\frac{hc}{E}\cdot\frac{\sqrt{2mE}}{h}=c\sqrt{\frac{2m}{E}}$ .

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Past questions and answers

4.An electron is accelerated from rest through a potential difference of VVV volt. If the de Broglie wavelength of the electron is 1.227×10−21.227\times10^{-2}1.227×10−2 nm, the potential difference is (a) 10310^3103 V (b) 10410^4104 V (c) 10 V (d) 10210^2102 V

5.Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is: (a) 527\frac{5}{27}275​ (b) 323\frac{3}{23}233​ (c) 729\frac{7}{29}297​ (d) 931\frac{9}{31}319​

6.The half-life of a radioactive isotope XXX is 20 years. It decays to another element YYY which is stable. The two elements XXX and YYY were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rock is estimated to be: (a) 40 years (b) 60 years (c) 80 years (d) 100 years

7.A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 u. The energy liberated per u is: (Given 1 u = 931 MeV) (a) 2.67 MeV (b) 26.7 MeV (c) 6.675 MeV (d) 13.35 MeV

12.The barrier potential of a p-n junction depends on: (a) type of semi conductor material (b) amount of doping (c) temperature Which one of the following is correct? (1) (b) only (2) (b) and (c) only (3) (a), (b) and (c) (4) (a) and (b) only

14.If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is: (a) 25 (b) 75 (c) 60 (d) 50

15.Hydrogen atom in ground state is excited by a monochromatic radiation of λ=975\lambda=975λ=975 Å. Number of spectral lines in the resulting spectrum emitted will be: (a) 2 (b) 6 (c) 10 (d) 3

17.When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is: (a) 1.0 eV (b) 1.3 eV (c) 1.5 eV (d) 0.65 eV

22.In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is: (a) 275\frac{27}{5}527​ (b) 527\frac{5}{27}275​ (c) 49\frac{4}{9}94​ (d) 94\frac{9}{4}49​

24.When an α\alphaα-particle of mass mmm moving with velocity vvv bombards on a heavy nucleus of charge ZeZeZe, its distance of closest approach from the nucleus depends on mmm as: (a) mmm (b) 1m\frac{1}{m}m1​ (c) 1m\frac{1}{\sqrt m}m​1​ (d) 1m2\frac{1}{m^2}m21​

33.The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is (a) 1 (b) 4 (c) 0.5 (d) 2

34.In a p-n junction diode, change in temperature due to heating (a) affects the overall V-I characteristics of p-n junction (b) affects only reverse resistance (c) does not affect resistance of p-n junction (d) affects only forward resistance

36.For a radioactive material, half-life is 10 minutes. If initially there are 600 nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is (a) 15 (b) 20 (c) 30 (d) 10

38.The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is (a) 1:−21:-21:−2 (b) 1:11:11:1 (c) 2:−12:-12:−1 (d) 1:−11:-11:−1

39.The total energy of an electron in an atom in an orbit is −3.4-3.4−3.4 eV. Its kinetic and potential energies are, respectively (a) −3.4-3.4−3.4 eV, −3.4-3.4−3.4 eV (b) −3.4-3.4−3.4 eV, −6.8-6.8−6.8 eV (c) 3.43.43.4 eV, −6.8-6.8−6.8 eV (d) 3.43.43.4 eV, 3.43.43.4 eV

40.α\alphaα-particle consists of (a) 2 protons and 2 neutrons only (b) 2 electrons, 2 protons and 2 neutrons (c) 2 electrons and 4 protons only (d) 2 protons only

41.Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled? (a) one-fourth (b) zero (c) doubled (d) four times

43.The increase in the width of the depletion region in a p-n junction diode is due to (a) both forward bias and reverse bias (b) increase in forward current (c) forward bias only (d) reverse bias only

48.A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the binding energy in the process is: (a) 804 MeV (b) 216 MeV (c) 0.9 MeV (d) 9.4 MeV

50.A half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be: (a) 23\frac{2}{3}32​ (b) 232\frac{2}{3\sqrt{2}}32​2​ (c) 12\frac{1}{2}21​ (d) 122\frac{1}{2\sqrt{2}}22​1​

51.Let T1T_1T1​ and T2T_2T2​ be the energy of an electron in the first and second excited states of hydrogen atom, respectively. According to the Bohr's model of an atom, the ratio T1:T2T_1 : T_2T1​:T2​ is: (a) 1 : 4 (b) 4 : 1 (c) 4 : 9 (d) 9 : 4

52.In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be: (a) Zero (b) 30 Hz (c) 60 Hz (d) 120 Hz

54.A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64. The ratio of radius of two daughter nuclei respectively is: (a) 1 : 1 (b) 4 : 5 (c) 5 : 4 (d) 25 : 16

55.The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of VVV volts is proportional to: (a) VVV (b) 1V\frac{1}{V}V1​ (c) 1V\frac{1}{\sqrt{V}}V​1​ (d) V\sqrt{V}V​

57.In hydrogen spectrum, the shortest wavelength in the Balmer series is λ\lambdaλ. The shortest wavelength in the Brackett series is: (a) 2λ2\lambda2λ (b) 4λ4\lambda4λ (c) 9λ9\lambda9λ (d) 16λ16\lambda16λ

59.The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to (116)th\left(\frac{1}{16}\right)^{th}(161​)th of its initial value? (a) 20 minutes (b) 40 minutes (c) 60 minutes (d) 80 minutes

60.A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a load resistance. Which of these components remove the ac ripple from the rectified output? (a) A centre-tapped transformer (b) p-n junction diodes (c) Capacitor (d) Load resistance

61.The radius of innermost orbit of hydrogen atom is 5.3×10−115.3 \times 10^{-11}5.3×10−11 m. What is the radius of third allowed orbit of hydrogen atom? (a) 0.53 A˚\mathrm{\AA}A˚ (b) 1.06 A˚\mathrm{\AA}A˚ (c) 1.59 A˚\mathrm{\AA}A˚ (d) 4.77 A˚\mathrm{\AA}A˚

67.De-Broglie wavelength of an electron orbiting in the n=2n = 2n=2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm): (a) 1.67 nm (b) 2.67 nm (c) 0.067 nm (d) 0.67 nm

An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227\times10^{-2}$ nm, the potential difference is
(a) $10^3$ V (b) $10^4$ V (c) 10 V (d) $10^2$ V

Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is:
(a) $\frac{5}{27}$ (b) $\frac{3}{23}$ (c) $\frac{7}{29}$ (d) $\frac{9}{31}$

The half-life of a radioactive isotope $X$ is 20 years. It decays to another element $Y$ which is stable. The two elements $X$ and $Y$ were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rock is estimated to be:
(a) 40 years (b) 60 years (c) 80 years (d) 100 years

A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 u. The energy liberated per u is: (Given 1 u = 931 MeV)
(a) 2.67 MeV (b) 26.7 MeV (c) 6.675 MeV (d) 13.35 MeV

The barrier potential of a p-n junction depends on:
(a) type of semi conductor material (b) amount of doping (c) temperature
Which one of the following is correct?
(1) (b) only (2) (b) and (c) only (3) (a), (b) and (c) (4) (a) and (b) only

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is:
(a) 25 (b) 75 (c) 60 (d) 50

Hydrogen atom in ground state is excited by a monochromatic radiation of $\lambda=975$ Å. Number of spectral lines in the resulting spectrum emitted will be:
(a) 2 (b) 6 (c) 10 (d) 3

When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is:
(a) 1.0 eV (b) 1.3 eV (c) 1.5 eV (d) 0.65 eV

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:
(a) $\frac{27}{5}$ (b) $\frac{5}{27}$ (c) $\frac{4}{9}$ (d) $\frac{9}{4}$

When an $\alpha$ -particle of mass $m$ moving with velocity $v$ bombards on a heavy nucleus of charge $Ze$ , its distance of closest approach from the nucleus depends on $m$ as:
(a) $m$ (b) $\frac{1}{m}$ (c) $\frac{1}{\sqrt m}$ (d) $\frac{1}{m^2}$

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is
(a) 1 (b) 4 (c) 0.5 (d) 2

In a p-n junction diode, change in temperature due to heating
(a) affects the overall V-I characteristics of p-n junction (b) affects only reverse resistance (c) does not affect resistance of p-n junction (d) affects only forward resistance

For a radioactive material, half-life is 10 minutes. If initially there are 600 nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
(a) 15 (b) 20 (c) 30 (d) 10

The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is
(a) $1:-2$ (b) $1:1$ (c) $2:-1$ (d) $1:-1$

The total energy of an electron in an atom in an orbit is $-3.4$ eV. Its kinetic and potential energies are, respectively
(a) $-3.4$ eV, $-3.4$ eV (b) $-3.4$ eV, $-6.8$ eV (c) $3.4$ eV, $-6.8$ eV (d) $3.4$ eV, $3.4$ eV

$\alpha$ -particle consists of
(a) 2 protons and 2 neutrons only (b) 2 electrons, 2 protons and 2 neutrons (c) 2 electrons and 4 protons only (d) 2 protons only

Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?
(a) one-fourth (b) zero (c) doubled (d) four times

The increase in the width of the depletion region in a p-n junction diode is due to
(a) both forward bias and reverse bias (b) increase in forward current (c) forward bias only (d) reverse bias only

A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the binding energy in the process is:
(a) 804 MeV (b) 216 MeV (c) 0.9 MeV (d) 9.4 MeV

A half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be:
(a) $\frac{2}{3}$ (b) $\frac{2}{3\sqrt{2}}$ (c) $\frac{1}{2}$ (d) $\frac{1}{2\sqrt{2}}$

Let $T_1$ and $T_2$ be the energy of an electron in the first and second excited states of hydrogen atom, respectively. According to the Bohr's model of an atom, the ratio $T_1 : T_2$ is:
(a) 1 : 4 (b) 4 : 1 (c) 4 : 9 (d) 9 : 4

In half wave rectification, if the input frequency is 60 Hz, then the output frequency would be:
(a) Zero (b) 30 Hz (c) 60 Hz (d) 120 Hz

A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64. The ratio of radius of two daughter nuclei respectively is:
(a) 1 : 1 (b) 4 : 5 (c) 5 : 4 (d) 25 : 16

The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to:
(a) $V$ (b) $\frac{1}{V}$ (c) $\frac{1}{\sqrt{V}}$ (d) $\sqrt{V}$

In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$ . The shortest wavelength in the Brackett series is:
(a) $2\lambda$ (b) $4\lambda$ (c) $9\lambda$ (d) $16\lambda$

The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $\left(\frac{1}{16}\right)^{th}$ of its initial value?
(a) 20 minutes (b) 40 minutes (c) 60 minutes (d) 80 minutes

A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a load resistance. Which of these components remove the ac ripple from the rectified output?
(a) A centre-tapped transformer (b) p-n junction diodes (c) Capacitor (d) Load resistance

The radius of innermost orbit of hydrogen atom is $5.3 \times 10^{-11}$ m. What is the radius of third allowed orbit of hydrogen atom?
(a) 0.53 $\mathrm{\AA}$ (b) 1.06 $\mathrm{\AA}$ (c) 1.59 $\mathrm{\AA}$ (d) 4.77 $\mathrm{\AA}$

De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius = 0.052 nm):
(a) 1.67 nm (b) 2.67 nm (c) 0.067 nm (d) 0.67 nm